FLUID PROPERTIES
Chapter 2
CE319F: Elementary Mechanics of Fluids
1
Fluid Properties
• Define “characteristics” of a specific fluid
•Properties expressed by basic “dimensions”
– length, mass (or force), time, temperature
• Dimensions quantified by basic “units”
We will consider systems of units, important fluid properties
(not all), and the dimensions associated with those properties.
2
Systeme International (SI)
•
•
•
•
Length = meters (m)
Mass = kilograms (kg)
Time = second (s)
Force = Newton (N)
– Force required to accelerate 1 kg @ 1 m/s2
– Acceleration due to gravity (g) = 9.81 m/s2
– Weight of 1 kg at earth’s surface = W = mg = 1 kg (9.81 m/s2) =
9.81 kg-m/s2 = 9.81 N
•
Temperature = Kelvin (oK)
– 273.15 oK = freezing point of water
– oK = 273.15 + oC
3
Système International (SI)
• Work and energy = Joule (J)
J = N*m = kg-m/s2 * m = kg-m2/s2
• Power = watt (W) = J/s
• SI prefixes:
G = giga = 109
M = mega = 106
k = kilo = 103
c = centi = 10-2
m = milli = 10-3
m = micro = 10-6
4
English (American) System
•
•
•
•
Length = foot (ft) = 0.3048 m
Mass = slug or lbm (1 slug = 32.2 lbm = 14.59 kg)
Time = second (s)
Force = pound-force (lbf)
– Force required to accelerate 1 slug @ 1 ft/s2
•
Temperature = (oF or oR)
– oRankine = oR = 460 + oF
• Work or energy = ft-lbf
• Power = ft-lbf/s
Banana Slug
Mascot of UC Santa Cruz
– 1 horsepower = 1 hp = 550 ft-lbf/s = 746 W
5
Density
• Mass per unit volume (e.g., @ 20 oC, 1 atm)
– Water
– Mercury
– Air
rwater = 1,000 kg/m3 (62.4 lbm/ft3)
rHg = 13,500 kg/m3
rair = 1.205 kg/m3
• Densities of gases = strong f (T,p) = compressible
• Densities of liquids are nearly constant
(incompressible) for constant temperature
• Specific volume = 1/density = volume/mass
6
Example: Textbook Problem 2.8
•
Estimate the mass of 1 mi3 of air in slugs and kgs.
Assume rair = 0.00237 slugs/ft3, the value at sea level for standard conditions
7
Example
•
A 5-L bottle of carbon tetrachloride is accidentally spilled onto a laboratory
floor. What is the mass of carbon tetrachloride that was spilled in lbm?
8
Specific Weight
g  rg
[ N / m3 ] or [lbf / ft 3 ]
• Weight per unit volume (e.g., @ 20 oC, 1 atm)
gwater
= (998 kg/m3)(9.807 m2/s)
= 9,790 N/m3
[= 62.4 lbf/ft3]
gair
= (1.205 kg/m3)(9.807 m2/s)
= 11.8 N/m3
[= 0.0752 lbf/ft3]
9
Specific Gravity
• Ratio of fluid density to density of water @
4 oC
rliquid
rliquid
SGliquid 

r water 1000 kg / m3
Water
Mercury
SGwater = 1
SGHg = 13.55
Note: SG is dimensionless and independent of system of units
10
Example
• The specific gravity of a fresh gasoline is 0.80. If the gasoline fills an
8 m3 tank on a transport truck, what is the weight of the gasoline in the
tank?
11
Ideal Gas Law (equation of state)
PV  nRuT
P = absolute (actual) pressure (Pa = N/m2)
V = volume (m3)
n = # moles
n
P  RuT
V
Ru = universal gas constant = 8.31 J/oK-mol
T = temperature (oK)
nM Ru
nM
P
T
RT  rRT
V M
V
R = gas-specific constant
R(air) = 287 J/kg-oK (show)
12
Example
• Calculate the volume occupied by 1 mol of any ideal gas at a
pressure of 1 atm (101,000 Pa) and temperature of 20 oC.
13
Example
• The molecular weight of air is approximately 29 g/mol. Use this
information to calculate the density of air near the earth’s surface
(pressure = 1 atm = 101,000 Pa) at 20 oC.
14
Example: Textbook Problem 2.4
• Given: Natural gas stored in a spherical tank
– Time 1: T1=10oC, p1=100 kPa
– Time 2: T2=10oC, p2=200 kPa
• Find: Ratio of mass at time 2 to that at time 1
• Note: Ideal gas law (p is absolute pressure)
15
Viscosity
16
Some Simple Flows
• Flow between a fixed and a moving plate
Fluid in contact with plate has same velocity as plate
(no slip condition)
u = x-direction component of velocity
y
Moving plate
u=V
V
B
u( y) 
V
y
B
Fluid
Fixed plate
x
17
u=0
Some Simple Flows
• Flow through a long, straight pipe
Fluid in contact with pipe wall has same velocity as wall
(no slip condition)
u = x-direction component of velocity
R
r
x
  r 2 
u (r )  V 1    
  R  
V
Fluid
18
Fluid Deformation
• Flow between a fixed and a moving plate
• Force causes plate to move with velocity V
and the fluid deforms continuously.
y
Moving plate
t0
u=V
t1 t2
Fluid
Fixed plate
x
u=0
19
Fluid Deformation
For viscous fluid, shear stress is proportional
to deformation rate of the fluid (rate of strain)
da

dt
y
dy
t
dL
da 
dy
dL
dt 
dV
dL
Moving plate
da
t+dt
dx
da dV

dt dy
u=V+dV
Fluid
Fixed plate
dV

dy
x
u=V
20
Viscosity
•
Proportionality constant = dynamic (absolute) viscosity
•
Newton’s Law of Viscosity
•
•
 m

Viscosity
m
Units
N / m2
dV / dy
m/s/m

dV
dy
V+d
v
V
N s
m2
m = 1x10-3 N-s/m2
•
Water (@ 20oC):
•
Air (@ 20oC): m = 1.8x10-5 N-s/m2
•
Kinematic viscosity  
m
r
Kinematic viscosity: m2/s
1 poise = 0.1 N-s/m2
1 centipoise = 10-2 poise = 10-3 N-s/m2
21
Shear in Different Fluids
• Shear-stress relations for different fluids
• Newtonian fluids: linear relationship
• Slope of line = coefficient of
proportionality) = “viscosity”
dV

dy
dV
 m
dy
Shear thinning fluids (ex): toothpaste, architectural coatings;
Shear thickening fluids = water w/ a lot of particles, e.g., sewage
sludge; Bingham fluid = like solid at small shear, then liquid at
greater shear, e.g., flexible plastics
22
Effect of Temperature
Gases:
greater T = greater interaction
between molecules = greater
viscosity.
Liquids:
greater T = lower cohesive forces
between molecules = viscosity
down.
23
24
Typical Viscosity Equations
3
Gas:
Liquid:
m  T  2 To  S
 
m To  T  S
m  Ce
b
T
T = Kelvin
S = Sutherland’s constant
Air = 111 oK
+/- 2% for T = 170 – 1900 oK
C and b = empirical constants
25
Flow between 2 plates
Force is same on top
and bottom
F1  1 A1   2 A2  F2
A1  A2
1   2
y
1  m
du
du
m
2
dy 1
dy 2
Thus, slope of velocity
profile is constant and
velocity profile is a st. line
Moving plate
u=V
V
B
u( y) 
V
y
B
Fluid
Fixed plate
Force acting
ON the plate
x
u=026
Flow between 2 plates
Shear stress anywhere
between plates
 m
du
V
m
dy
B
m  0.1 N  s / m 2 ( SAE 30 @ 38o C )   (0.1 N  s / m 2 )( 3 m / s )
0.02 m
V  3 m/s
B  0.02 m
y
 15 N / m 2
Moving plate
V
B
u( y) 
V
y
B
u=V


Fixed plate
Shear
on fluid
x
27
u=0
Flow between 2 plates
• 2 different coordinate systems
B
r
x
  r 2 
u (r )  V 1    
  B  
V
u ( y )  C  y  B  y 
y
x
28
Example: Textbook Problem 2.33
Suppose that glycerin is flowing (T = 20 oC) and that the pressure
gradient dp/dx = -1.6 kN/m3. What are the velocity and shear stress at a
distance of 12 mm from the wall if the space B between the walls is 5.0
cm? What are the shear stress and velocity at the wall? The velocity
distribution for viscous flow between stationary plates is

1 dp
u  
By
2m dx

y2
29

30
Example: Textbook Problem 2.34
A laminar flow occurs between two horizontal parallel plates under a
pressure gradient dp/ds (p decreases in the positive s direction). The upper
plate moves left (negative) at velocity ut. The expression for local velocity
is shown below. Is the magnitude of the shear stress greater at the moving
plate (y = H) of at the stationary plate (y = 0)?
u
 1 dp
y
Hy  y 2  ut
2m ds
H


31
32
Elasticity (Compressibility)
• If pressure acting on mass of fluid increases: fluid contracts
• If pressure acting on mass of fluid decreases: fluid expands
• Elasticity relates to amount of deformation for a given
change in pressure
dV  Vdp
1
dV  Vdp
Ev
Ev  
Ev = bulk modulus of elasticity
dp
dp

dV
dr
V
r
Small dV/V = large modulus of elasticity
How does second part of
equation come about?
33
Example: Textbook Problem 2.45
• Given: Pressure of 2 MPa is
applied to a mass of water that
initially filled 1000-cm3
(1 liter) volume.
•
• Find: Volume after the
pressure is applied.
• Ev = 2.2x109 Pa (Table A.5)
34
Example
• Based on the definition of Ev and the equation of state, derive an
equation for the modulus of elasticity of an ideal gas.
35
Surface Tension
• Below surface, forces act equal in all
directions
• At surface, some forces are missing, pulls
molecules down and together, like
membrane exerting tension on the surface
• Pressure increase is balanced by surface
tension, s
•
surface tension = magnitude of
tension/length
Interface
water
air
Net force
inward
No net force
 s = 0.073 N/m (water @ 20oC)
36
Surface Tension
• Liquids have cohesion and adhesion, both involving molecular
interactions
– Cohesion: enables liquid to resist tensile stress
– Adhesion: enables liquid to adhere to other bodies
• Capillarity = property of exerting forces on fluids by fine tubes
or porous media
–
–
–
–
–
due to cohesion and adhesion
If adhesion > cohesion, liquid wets solid surfaces at rises
If adhesion < cohesion, liquid surface depresses at pt of contact
water rises in glass tube (angle = 0o)
mercury depresses in glass tube (angle = 130-140o)
• See attached information
37
Example: Capillary Rise
• Given: Water @ 20oC, d = 1.6 mm
• Find: Height of water
Fs
W
38
Example: Textbook Problem 2.51
Find: Maximum capillary
rise between two vertical
glass plates 1 mm apart.
q
s
s
h
t
39
Examples of Surface Tension
40
Example: Textbook Problem 2.48
Given: Spherical soap bubble, inside
radius r, film thickness t, and surface
tension s.
Find: Formula for pressure in the
bubble relative to that outside.
Pressure for a bubble with a 4-mm
radius?
Should be soap bubble
41
Vapor Pressure (Pvp)
• Vapor pressure of a pure liquid = equilibrium partial pressure of the gas
molecules of that species above a flat surface of the pure liquid
– Concept on board
– Very strong function of temperature (Pvp up as T up)
– Very important parameter of liquids (highly variable – see attached page)
• When vapor pressure exceeds total air pressure applied at surface, the liquid
will boil.
• Pressure at which a liquid will boil for a given temperature
– At 10 oC, vapor pressure of water = 0.012 atm = 1200 Pa
– If reduce pressure to this value can get boiling of water (can lead to “cavitation”)
• If Pvp > 1 atm compound = gas
• If Pvp < 1 atm compound = liquid or solid
42
Example
• The vapor pressure of naphthalene at 25 oC is 10.6 Pa. What is the
corresponding mass concentration of naphthalene in mg/m3? (Hint:
you can treat naphthalene vapor as an ideal gas).
43
Vapor Pressure (Pvp) - continued
Vapor Press. vs. Temp.
120
Vapro Pressure (kPa)
100
80
60
40
20
0
0
10
20
30
40
50
60
70
80
90
100
Temperature (oC)
Vapor pressure of water (and other liquids) is a strong function of temperature.
44
Vapor Pressure (Pvp) - continued
Pvp,H2O = Pexp(13.3185a – 1.9760a2 – 0.6445a3 – 0.1299a4)
P = 101,325 Pa
a = 1 – (373.15/T)
T = oK
valid to +/- 0.1% accuracy for T in range of -50 to 140 oC
RH  100% x
PH 2O
Pvp , H 2O
Equation for relative humidity of air = percentage to which air is “saturated” with water vapor.
What is affect of RH on drying of building materials, and why? Implications?
45
Example: Relative Humidity
The relative humidity of air in a room is 80% at 25 oC.
(a) What is the concentration of water vapor in air on a volume percent
basis?
(b) If the air contacts a cold surface, water may condense (see effects on
attached page). What temperature is required to cause water
condensation?
46
47
Saturation Vapor Pressure
4500
4000
3500
Pvp (Pa)
3000
2500
2000
1500
1000
500
0
0
5
10
15
20
degrees C
25
30
35
48
49